# Ow long does it take for the charge to build up to 9.0 µC? RC circuits

1. ### mr_coffee

Hello everyone, I almost got all the parts to this problem:
In an RC series circuit, = 11.0 V, R = 1.10 M, and C = 1.80 µF.

(a) Calculate the time constant.
1.98 s
(b) Find the maximum charge that will appear on the capacitor during charging.
19.8 µC
(c) How long does it take for the charge to build up to 9.0 µC?
s

I tried the following:
q(t) = Q[1-e^(-t/RC)]
q(t) = 9.0e-6[1-e^(-1.98)] = 7.75e-6s, which is wrong;
Since i know t = RC, and i found RC = 1.98, would it be
q(t) = 9.0e-6[1-e^(-1)] ?

I just tried this and its also wrong.

2. ### daniel_i_l

866
In your equation q(t) should equal 9.0mc since that is the charge that has built up after time t. and Q which is the total charge should be 19.8mc, then just solve for time.

3. ### mr_coffee

thanks! i'm stuck on how i'm suppose to solve for t...
i have -(19.8E-6)(e^(-t/1.98));
do i take the natural log of both sides or somthing? but u can't take the natural log of a negative number!

4. ### daniel_i_l

866
you know that:
9.0E-6 = 19.8E-6[1-e^(-t/1.98)] = 19.8E-6 - 19.8E-6*e^(-t/1.98) so:

-10.8E-6 = -19.8E-6*e^(-t/1.98) /-1
10.8E-6 = 19.8E-6*e^(-t/1.98)
now you can take the natural log of both sides

5. ### mr_coffee

Hey thanks for the responce!
I took the ln of both sides and got:
-11.44 = (-t/1.98)*ln(19.8E-6);
I end up with a negative number of -10.8, did I do this wrong? Sorry its been awhile since I did stuff with ln

6. ### big man

254
you didn't take the natural logarithm of each side in the correct manner.
First divide through by 19.8*10^-6 to get rid of that term on the RHS of the expression.

So you would then have:

0.54 = e^(-t/1.98)

Take the natural log of that and you have:

ln(0.54) = -t/1.98
This will give you a positive number

7. ### mr_coffee

Thanks!! But i submitted that as my answer: t = .311 and it was also incorrect!! hm...

8. ### mr_coffee

ohh my bad hah thanks big man, it worked great!!!